Henry's little brother has $8$ identical stickers and $4$ sheets of paper, each a different color. He puts all the stickers on the pieces of paper. How many ways are there for him to do this, if only the number of stickers on each sheet of paper matters?
Answer: Ignoring the different colors of paper, we can put the stickers on the sheets of paper in the following groups: \begin{align*}
& (8,0,0,0) \\
& (7,1,0,0) \\
& (6,2,0,0) \\
& (6,1,1,0) \\
& (5,3,0,0) \\
& (5,2,1,0) \\
& (5,1,1,1) \\
& (4,4,0,0) \\
& (4,3,1,0) \\
& (4,2,2,0) \\
& (4,2,1,1) \\
& (3,3,2,0) \\
& (3,3,1,1) \\
& (3,2,2,1) \\
& (2,2,2,2).
\end{align*}For each of these combinations, we will list how many distinct ways there are to put the groups of stickers on the different sheets of paper.

$\bullet$  For $(8,0,0,0),$ there are $\dfrac{4!}{3!}=4$ ways.

$\bullet$  For $(7,1,0,0),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(6,2,0,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(6,1,1,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(5,3,0,0),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(5,2,1,0),$ there are $4!=24$ ways.

$\bullet$  For $(5,1,1,1),$ there are $\dfrac{4!}{3!}=4$ ways.

$\bullet$  For $(4,4,0,0),$ we have $\dfrac{4!}{2!2!}=6$ ways.

$\bullet$  For $(4,3,1,0),$ there are $4!=24$ ways.

$\bullet$  For $(4,2,2,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(4,2,1,1),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(3,3,2,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(3,3,1,1),$ there are $\dfrac{4!}{2!2!}=6$ ways.

$\bullet$  For $(3,2,2,1),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$  For $(2,2,2,2),$ there are $\dfrac{4!}{4!}=1$ way.

In total, there are $$4+12+12+12+12+24+4+6+24+12+12+12+6+12+1=\boxed{165}$$ways for Henry's brother to put the stickers on the sheets of paper.

Notice that this answer equals $\binom{11}{3}.$ Is this a coincidence?